Thursday, February 10, 2011

What proportion of total BOD of wastewater would be utilized in five days with k1’ values of 0.05, 0.1, 0.15, 0.20 and 0.25?

The biochemical oxygen demand or BOD is simply the amount of oxygen required for biological decay of organic material present in the sample. Mathematically it is expressed as:


`BOD_t = BOD_u (1 - e^(-kt))`


where BODu is the ultimate BOD or the total amount of oxygen needed for complete decay, while BODt is the amount of oxygen needed in time t. The rate constant, k,  is given in the units of per day. The equation can also be written as:


`BOD_t = BOD_u (1- 10^(-k't))`


Depending on the author, k and k' may be represented by k1 and k2 or k1 and k1', etc. The only difference between these values is where we use base e or base 10 in the equation. 


We can rearrange the equation as following to determine the proportion of total BOD expressed in a given time:


`(BOD_t) / (BOD_u) = 1 - e^-(kt) = 1- 10^-(k't)`


Since it is not clear from the question whether we have to use base "e" or base "10", I am using base 10 for the calculations.


The time t = 5 days


For k1' = 0.05, `(BOD_5)/(BOD_u) = 1- 10^-(0.05 xx 5) = 0.44`


For k1' = 0.1, `(BOD_5)/(BOD_u) = 1-10^-(0.1xx5) = 0.68`


Similarly for k1' values of 0.15, 0.2 and 0.25, the proportion will be 0.82, 0.9 and 0.94, respectively. 


If you need to use base "e," then the equation involving e (1-e^-kt) can be used. The answers in that case would be 0.22, 0.39, 0.53, 0.63 and 0.71, respectively for k1' values of 0.05, 0.1, 0.15, 0.2 and 0.25.

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